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Take a square paper of size S X S units, and thickness of paper 1 unit.

How many half folds should we do to get form a cube out it?

Consider infinite foldable paper.

Firstly there must be an even number of folds for the folded paper to remain square


Consider the volume of the paper which is $S times S times 1 = S^2$


Therefore the resultant cube side is $sqrt[3]S^2$


After $N$ pairs of folds the side length is $S / (2^N)$


So $S / (2^N) = sqrt[3]S^2$


So $S = 2^3N$


So $N = log S / (3 times log 2) $

Or $N = log S / log 8 $


Now suppose $ S = 512$ thickness $T = 1$

The computation gives $ N = 3 $ pairs of folds


Worked example:

After 1st pair of folds $S = 256$ with $T = 4$

After 2nd pair of folds $S = 128$ with $T = 16$

After 3rd pair of folds $S = 64$ with $T = 64$ which is a cube


The question asks how many half folds?

Answer:

Half folds = $ 2 times log S / log 8 $


Obviously the paper can only be folded thus if $N$ is an integer

Consider what happens by doing $2$ folds, one in each direction.

You now have a square that is half the size, but $4$ times the thickness.

So after $2k$ folds

the square has edge length $S/2^k$ and thickness $4^k$.

For this to be a cube we need:

$$fracS2^k = 4^k\ S=8^k \ k = log_8S = fraclog Slog 8$$

So the number of folds we need is

$$2k = frac2log Slog 8$$
It will only be a cube if this result is an even whole number, i.e. if S is a power of $8$.

In reality, this will only work if you cut the paper in half and stack the pieces, instead of folding the paper. I have ignored the amount of paper connecting the different layers of the folded cube, which is quite substantial, and the round folded edges also keep it from being in the shape of a cube.

I think it should be...

$N = 2 log_2sqrt[leftroot-2uproot23]S$, where $N$ is the number of folds.

As a caveat, $S$ can only be $2^3x$, where $x in mathbbN$.


For example:
$S=4096$ ($=2^12)$
$sqrt[leftroot-2uproot23]S = 16$; $log_216 = 4$; $4 times 2=8$ folds.
$4096 div 2^4$ (I fold each side $4$ times) $= 256$.

Thickness is $2^8$ ($8$ folds) = $256$.

It is simple.

The volume of paper is $S^2$.

So the side of the cube we want is $sqrt[3]S^2 = S^frac23$.

Each fold doubles the thickness. It starts with $1$. After $N$ folds the thickness is $2^N$.

To get the correct thickness for the cube we need to have $2^N = S^frac23$.

Taking $log_2$ on both sides we get $N = frac23 log_2 S$.

Of course, it only works if N is an even integer.